Import Question JSON

Current Question (ID: 10825)

Question:
$\text{The } C - H \text{ bond distance is longer in -}$
Options:
  • 1. $\text{C}_2\text{H}_2$
  • 2. $\text{C}_2\text{H}_4$
  • 3. $\text{C}_2\text{H}_6$ (Correct)
  • 4. $\text{C}_2\text{H}_2\text{Br}_2$
Solution:
$\text{Hint: As the size of the hybrid orbital increases, the C-H bond length also increases.}$ $\text{The structure of the compounds are as follows:}$ $\text{H-C}\equiv\text{C-H (sp carbon), H}_2\text{C=CH}_2\text{ (sp}^2\text{ carbon), H}_3\text{C-CH}_3\text{ (sp}^3\text{ carbon), and Br-HC=CH-Br (sp}^2\text{ carbon)}$ $\text{The size of the hybrid orbital of carbon of } C - H \text{ bond decreases in the order: sp}^3 > \text{sp}^2 > \text{sp. Therefore, } C - H \text{ bond distance decreases in the same order. Thus } C - H \text{ bond distance in C}_2\text{H}_6 \text{ is the longest.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}